Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 5 Numerical Methods Ex 5.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 5 Numerical Methods Ex 5.1

Question 1.

Evaluate Δ (log ax).

Solution:

Δ log (ax) = log (ax + h) – log ax

= log [ \(\frac { ax+h }{ax}\) ] = log[\(\frac { ax }{ax}\) + \(\frac { h }{ax}\)]

= log [1 + \(\frac { h }{ax}\)]

Question 2.

If y = x³ – x² + x -1 calculate the values of y for x= 0, 1, 2, 3, 4, 5 and form the forward differences table.

Solution:

Given y = x³ – x² + x – 1

when x = 0 y = -1

when x = 1

y = 1 – 1 + 1 – 1 = 0

when x = 2

y = 8 – 4 + 2 – 1 = 5

for x = 0, 1, 2, 3, 4, 5

when x = 3

y = 27 – 9 + 3 – 1 = 20

when x = 4

y = 64 – 16 + 4 – 1 = 51

when x = 5

y = 125 – 25 + 5 – 1 = 104

Question 3.

If h = 1 then prove that (E^{-1} Δ)x³ = 3x² – 3x + 1

Solution:

h = 1 To prove (E^{-1} ∆) x3 = 3×2 – 3x + 1

L.H.S = (E^{-1} ∆) x^{3} = E^{-1} (∆x^{3})

= E^{-1}[(x + h)^{3} – x^{3}]

= E^{-1}( x + h)^{3} – E^{-1}(x^{3})

= (x – h + h)^{3} – (x – h)^{3}

= x^{3} – (x – h)^{3}

But given h = 1

So(E^{-1} ∆) x^{3} = x^{3} – (x – 1)^{3}

= x^{3} – [x^{3} – 3x^{2} + 3x – 1]

= 3x^{2} – 3x + 1

= RHS

Question 4.

If f(x) = x² + 3x than show that Δf(x) = 2x + 4

Solution:

Given f(x) = x³ + 3x; h = 1

Δf(x) = f (x + h) – f(x)

= (x + 1)² + 3 (x + 1) – (x² + 3x)

= x² + 2x + 1 + 3x + 3 – x² + 3x

= 2x + 4

∴ Δf(x) = 2x + 4

Question 5.

Evaluate Δ [ \(\frac { 1 }{(x+1)+(x+2)}\) ] by taking ‘1’ as the interval of differencing

Solution:

Question 6.

Find the missing entry in the following table

Solution:

Since only four values of f(x) are given the polynomial which fits the data is of degree three. Hence fourth differences are zeros.

(ie) Δ^{4} y_{0} = 0

(E – 1)^{4} y_{0} = 0

(E^{4} – 4E³ + 6E² – 4E + 1) y_{0} = 0

E^{4}y_{0} – 4E³ y_{0} + 6E²y_{0} – 4E y_{0} + 1y_{0} = 0

y_{4} – 4y_{3} + 6y_{2} – 4y_{1}+ y_{0} = o

81 – 4y_{3} + 6(9) – 4(3) + 1 = 0

81 – 4y_{3} + 54 – 12 + 1 = 0

136 – 12 – 4y_{3} = 0

4y_{3} = 124

y_{3} = \(\frac { 124 }{4}\)

∴ y_{3} = 31

Question 7.

Following are the population of a district

Find the population of the year 1911

Solution:

Since only five values of fix) are given, the polynomial which fits the data is of degree four. Hence fifth differences are zeros.

(ie) Δ^{5} y_{0} = 0

(E – 1)^{5} y_{0} = 0

(E^{5} – 5E^{4} + 10E³ – 10E² + 5E – 1) y_{0} = 0

E^{5}y_{0} – 5E^{4}y_{0} + 10E³y_{0} – 10E²y_{0} + 5E y_{0} – y_{0} = 0

y_{5} – 5y_{4} + 10y_{3} – 10y_{2} + 5y_{1} – y_{0} = 0

501 – 5 (467) + 10(y_{3}) -10 (421) + 5 (391) – 363 = 0

2456 – 6908 + 10y_{3} = 0

-4452 + 10y_{3} = 0 ⇒ 10y_{3} = 4452

y = \(\frac { 4452 }{10}\) = 445.2

The population of the year 1911 is 445.2 thousands

Question 8.

Find the missing entries from the following.

Solution:

Since four values of f(x) are given we assume the polynomial of degree three. Hence fourth order differences are zeros.

(ie) Δ^{4}y_{0} = 0 (ie) (E – 1)^{4} y_{k} = 0

(E^{4} – 4E³ + 6E² – 4E + 1) y_{k} = 0 ……… (1)

Put k = 0 in (1)

(E^{4} – 4E³ + 6E² – 4E + 1) y_{0} = 0

E^{4}y_{0} – 4E³y_{0} + 6E³y_{0} – 4Ey_{0} + y_{0} = 0

y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0

y_{4} – 4 (15) + 6(8) – 4y_{1} + 0 = 0

y_{4} – 4y_{1} = 12 …….. (2)

Put k = 1 in eqn (1)

(E^{4} – 4E³ + 6E² – 4E + 1) y_{1} = 0

y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0

35 – 4 (y_{4}) + 6(15) – 4(8) + y_{1} = 0

35 – 4y_{4} + 90 – 32 + y_{1} = 0

-4y_{4} + y_{1} + 125 – 32 = 0

-4y_{4} + y_{1} = -93 ………. (3)

Solving eqn (2) & (3)

Substitute y = 3 in eqn (2)

y_{4} – 4(3) = 12

y_{4} – 12 = 12

y_{4} = 12 + 12

∴ y_{4} = 24

The required two missing entries are 3 and 24.